4
$\begingroup$

There is an exam problem I'm having trouble with, it is as follows:

Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.

Comparing to the general equation for conic sections I see that this is a rotated ellipse. I assume that the solution would be to multiply a rotation matrix (to align the ellipse's axes to the $x$ and $y$ axes) and a translation matrix (to bring the ellipse's center to the origin), and then somehow apply the result to the ellipse.

How would I go about doing this? How do I get the coordinates of the ellipse's center and the angle of its rotation to construct the matrices, and then how would I apply the transformation to the ellipse itself?

$\endgroup$
1
  • $\begingroup$ Have a look at this. $\endgroup$ Commented Aug 3, 2017 at 14:22

4 Answers 4

3
$\begingroup$

To keep it understandable albeit inelegant I'll do the passage in two steps

First the translation

$x^2 - xy + y^2 - 3y -1 = 0$

we look for a new centre $(h;\;k)$ so we substitute $x=x'+h;\;y=y'+k$

$-(h+x) (k+y)+(h+x)^2+(k+y)^2-3 (k+y)-1=0$

$x'^2-x' y'+y^2+x' (2 h-k)+y' (-h+2 k-3) +h^2-h k+k^2-3 k-1=0$

To have no first degree terms we put $2h-k=0;\;-h+2 k-3=0$ which gives

$h=1;\;k=2$ and we plug these values in the previous equation

$x'^2 - x' y' + y'^2=4$

Now we want to get rid of the $x'y'$ term. To do so we have to rotate the axis using these equations

$\begin{aligned}x'&=X\cos \theta -Y\sin \theta \\y'&=X\sin \theta +Y\cos \theta \end{aligned}$

$(X \sin \theta+Y \cos \theta)^2-(X \cos \theta-Y \sin \theta) (X \sin \theta+Y \cos \theta)+(X \cos \theta-Y \sin \theta)^2=4$

collecting terms

$X^2 \left(\sin ^2\theta+\cos ^2\theta-\sin \theta \cos \theta\right)+X Y \left(\sin ^2\theta-\cos ^2\theta\right)+Y^2 \left(\sin ^2\theta+\cos ^2\theta+\sin \theta \cos \theta\right)=4$

as we want the term $XY$ off we set $\sin ^2\theta-\cos ^2\theta=0$

$\tan^2\theta=1\to \theta=\pm \dfrac{\pi}{4}$

if we want the major axis of the ellipse to be horizontal we choose $\theta=\dfrac{\pi}{4}$ and substitute this value in the last equation

$\dfrac{X^2}{8}+\dfrac{3 Y^2}{8}=1$

the equations of the roto-translation altogether are

$\begin{aligned}x&=X\cos \frac{\pi}{4}-Y\sin \frac{\pi}{4} +1\\y&=X\sin \frac{\pi}{4}+Y\cos \frac{\pi}{4} +2\end{aligned}$

or

$\begin{aligned}x&=\frac{\sqrt 2}{2}(X-Y) +1\\y&=\frac{\sqrt 2}{2}(X+Y) +2\end{aligned}$

hope this helps

$\endgroup$
2
  • $\begingroup$ Thank you very much! To me this is the most understandable and most elegant answer, and very doable with pen and paper in the limited amount of time. $\endgroup$ Commented Aug 4, 2017 at 11:43
  • $\begingroup$ @Raffaele, I went through wikipedia on canonical form, it didnt make much sense what canonical form actually is, so looking at your answer, is it the form where two dimensions are seperated instead of a single line equation? $\endgroup$ Commented Dec 1, 2020 at 12:45
3
$\begingroup$

First center the equation by definting the function $f(x,y) = x^2 - xy + y^2 - 3y - 1 = 0$ and solving the following system of equations for $(x,y)$

$$\left. \begin{aligned} \frac{\partial f(x,y)}{\partial x} & = 2x-y = 0 \\ \frac{\partial f(x,y)}{\partial y} & = -x+2y-3 = 0 \\ \end{aligned} \right\} \pmatrix{x & y} = \pmatrix{1 & 2} $$

So now we reset the coordinates to $\pmatrix{x&y} \rightarrow \pmatrix{x+1&y+2}$

The new equation is $g(x,y) = x^2-x y + y^2 -4 = 0$

Now to re-orient the conic. Use $\pmatrix{x&y} \rightarrow \pmatrix{x \cos \theta - y \sin \theta & x \sin\theta + y \cos \theta}$ and set the coefficient of $x y$ equal to zero.

$$ g = \left( 1- \frac{\sin 2 \theta}{2} \right) x^2 + \left( 1 + \frac{\sin 2 \theta}{2} \right) y^2 +2 \left( -\frac{\cos 2\theta}{2} \right) x y -4 =0 $$

$$ \left. -\frac{1}{2}\cos(2 \theta)=0 \right\} \theta =\pm \frac{\pi}{4} $$

$$\boxed{ d(x,y) = \frac{1}{2} x^2 + \frac{3}{2} y^2 -4 =0 }$$

$\endgroup$
1
  • 1
    $\begingroup$ This is a very elegant answer, though perhaps a bit of insight can be added. Why should we calculate those partial derivatives? One way to think about this implicitly-defined set is as the intersection of the $xy$ plane ($z=0$) with the graph of a function $z=f(x,y)$. Thinking about the paraboloid as an example, one can notice that at its center is at a critical point whose tangent plane is "flat", or whose partial derivatives are both zero. We can use this tangent plane property to go about finding that special point, the "center" of the object, and therefore set the partials to 0. $\endgroup$
    – Menachem
    Commented Nov 3, 2021 at 15:10
2
$\begingroup$

You have to find an orthonormal basis of eigenvectors associated to the matrix of the quadratic part of the equation: $$Q=\begin{bmatrix} 1& -\dfrac12\\ -\dfrac12&1 \end{bmatrix}$$ The characteristic polynomial is $\;\chi_Q(\lambda)=(\lambda-1)^2-\dfrac14$, whence the eigenvalues $\lambda_1=\dfrac12$, $\;\lambda_2=\dfrac32$, and the corresponding eigenvectors $(1,1)$ and $(-1,1)$.

Normalise these vectors to obtain an orthonormal basis, and the change of basis matrix $$u=\biggl(\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad v=\biggl(-\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad P=\begin{bmatrix} \dfrac1{\sqrt2}& -\dfrac1{\sqrt2}\\ \dfrac1{\sqrt2}&\dfrac1{\sqrt2} \end{bmatrix}.$$ Usibg this matrix, we obtain the equation of the conic in the new coordinate system $(x',y')$: $$\frac12x'^2+\dfrac32y'^2-\dfrac3{\sqrt2}(x'+y')=1$$ Let's centre this equation rewriting it as \begin{align} &\phantom{\iff}\;\frac12\bigl(x'^2-3\sqrt 2x'\bigr)+\frac32\bigl(y'^2-\sqrt2 y')=1\\ &\iff\;\frac12\biggl(x'-\frac{3\sqrt 2}2\biggr)^2+\frac32\biggl(y'-\frac{\sqrt 2}2\biggr)^2=1+\frac94+\frac34=4 \end{align} So, setting \begin{cases} X=x'-\dfrac{3\sqrt2}2=\dfrac{\sqrt2}2(x+y-3),\\ Y=y'-\dfrac{\sqrt 2}2=\dfrac{\sqrt2}2(-x+y-1), \end{cases} we obtain the equation in canonical form: $$\frac{X^2}8+\frac{3Y^2}8=1.$$

$\endgroup$
0
$\begingroup$

$$\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2-3y-1=0$$ or

$$\left(x-\frac{y}{2}\right)^2+3\left(\frac{y}{2}-1\right)^2=4$$ or $$\frac{\left(x-\frac{y}{2}\right)^2}{4}+\frac{\left(\frac{y}{2}-1\right)^2}{\frac{4}{3}}=1.$$ Now, let $x'=x-\frac{y}{2}$ and $y'=\frac{y}{2}-1$...

$\endgroup$
2
  • $\begingroup$ But is it an isometry? $\endgroup$
    – Bernard
    Commented Aug 3, 2017 at 17:24
  • $\begingroup$ @Bernard No, of course. But we can a bit of change something. See your solution. $\endgroup$ Commented Aug 3, 2017 at 17:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .